Mixing
Finding the length of the side of a triangle, given the lengths of the other two sides.
$begingroup$
$ABC$ is a triangle with $AB=86$ and $AC=97$. A circle is drawn with $AB$ as radius which cuts $BC$ at $B$ and $X$ such that length of $BX$ and $XC$ is an integer. Find the length of $BC$.
Please help me out with this question.
geometry
edited Jan 2 at 16:12
amWhy
192k28225439
asked Jan 2 at 13:33
Kartik AggarwalKartik Aggarwal
41
$endgroup$
add a comment |
$begingroup$
$ABC$ is a triangle with $AB=86$ and $AC=97$. A circle is drawn with $AB$ as radius which cuts $BC$ at $B$ and $X$ such that length of $BX$ and $XC$ is an integer. Find the length of $BC$.
Please help me out with this question.
geometry
edited Jan 2 at 16:12
amWhy
192k28225439
asked Jan 2 at 13:33
Kartik AggarwalKartik Aggarwal
41
$endgroup$
$begingroup$
Maybe you can draw a picture and see what you you can deduce after that.
$endgroup$
– Matti P.
Jan 2 at 13:36
add a comment |
$begingroup$
$ABC$ is a triangle with $AB=86$ and $AC=97$. A circle is drawn with $AB$ as radius which cuts $BC$ at $B$ and $X$ such that length of $BX$ and $XC$ is an integer. Find the length of $BC$.
Please help me out with this question.
geometry
edited Jan 2 at 16:12
amWhy
192k28225439
asked Jan 2 at 13:33
Kartik AggarwalKartik Aggarwal
41
$endgroup$
$ABC$ is a triangle with $AB=86$ and $AC=97$. A circle is drawn with $AB$ as radius which cuts $BC$ at $B$ and $X$ such that length of $BX$ and $XC$ is an integer. Find the length of $BC$.
Please help me out with this question.
geometry
geometry
edited Jan 2 at 16:12
amWhy
192k28225439
asked Jan 2 at 13:33
Kartik AggarwalKartik Aggarwal
41
edited Jan 2 at 16:12
amWhy
192k28225439
asked Jan 2 at 13:33
Kartik AggarwalKartik Aggarwal
41
edited Jan 2 at 16:12
amWhy
192k28225439
edited Jan 2 at 16:12
amWhy
192k28225439
edited Jan 2 at 16:12
amWhy
192k28225439
192k28225439
asked Jan 2 at 13:33
Kartik AggarwalKartik Aggarwal
41
asked Jan 2 at 13:33
Kartik AggarwalKartik Aggarwal
41
asked Jan 2 at 13:33
Kartik AggarwalKartik Aggarwal
41
41
$begingroup$
Maybe you can draw a picture and see what you you can deduce after that.
$endgroup$
– Matti P.
Jan 2 at 13:36
add a comment |
$begingroup$
Maybe you can draw a picture and see what you you can deduce after that.
$endgroup$
– Matti P.
Jan 2 at 13:36
$begingroup$
Maybe you can draw a picture and see what you you can deduce after that.
$endgroup$
– Matti P.
Jan 2 at 13:36
$begingroup$
Maybe you can draw a picture and see what you you can deduce after that.
$endgroup$
– Matti P.
Jan 2 at 13:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that $ABX$ and $AXC$ will be right-angled triangles, with $angle AXB$ and $angle AXC$ being $90$ degree. Let $BX=x, CX=y$ and $AX=z$. Then,
$$y^2+z^2=97^2\and\x^2+z^2=86^2$$
Subtract them to get
$$(y+x)(y-x)=97^2-86^2=11cdot183=1cdot3cdot11cdot61$$
$y-x$ can attain the values $1, 3, 11$ and $33$. It can't attain others because it can't be greater than $y+x$. If $y-x=1$, $y+x=2013$. But, $AB+AC$ should be more than $BC$, which isn't the case. So, $y-x=1$ is rejected. Similarly, $3$ and $11$ won't be acceptable either. Hence, $y-x=33implies BC=x+y=61$.
answered Jan 2 at 15:10
Ankit KumarAnkit Kumar
1,389219
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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votes
$begingroup$
Note that $ABX$ and $AXC$ will be right-angled triangles, with $angle AXB$ and $angle AXC$ being $90$ degree. Let $BX=x, CX=y$ and $AX=z$. Then,
$$y^2+z^2=97^2\and\x^2+z^2=86^2$$
Subtract them to get
$$(y+x)(y-x)=97^2-86^2=11cdot183=1cdot3cdot11cdot61$$
$y-x$ can attain the values $1, 3, 11$ and $33$. It can't attain others because it can't be greater than $y+x$. If $y-x=1$, $y+x=2013$. But, $AB+AC$ should be more than $BC$, which isn't the case. So, $y-x=1$ is rejected. Similarly, $3$ and $11$ won't be acceptable either. Hence, $y-x=33implies BC=x+y=61$.
answered Jan 2 at 15:10
Ankit KumarAnkit Kumar
1,389219
$endgroup$
add a comment |
$begingroup$
Note that $ABX$ and $AXC$ will be right-angled triangles, with $angle AXB$ and $angle AXC$ being $90$ degree. Let $BX=x, CX=y$ and $AX=z$. Then,
$$y^2+z^2=97^2\and\x^2+z^2=86^2$$
Subtract them to get
$$(y+x)(y-x)=97^2-86^2=11cdot183=1cdot3cdot11cdot61$$
$y-x$ can attain the values $1, 3, 11$ and $33$. It can't attain others because it can't be greater than $y+x$. If $y-x=1$, $y+x=2013$. But, $AB+AC$ should be more than $BC$, which isn't the case. So, $y-x=1$ is rejected. Similarly, $3$ and $11$ won't be acceptable either. Hence, $y-x=33implies BC=x+y=61$.
answered Jan 2 at 15:10
Ankit KumarAnkit Kumar
1,389219
$endgroup$
add a comment |
$begingroup$
Note that $ABX$ and $AXC$ will be right-angled triangles, with $angle AXB$ and $angle AXC$ being $90$ degree. Let $BX=x, CX=y$ and $AX=z$. Then,
$$y^2+z^2=97^2\and\x^2+z^2=86^2$$
Subtract them to get
$$(y+x)(y-x)=97^2-86^2=11cdot183=1cdot3cdot11cdot61$$
$y-x$ can attain the values $1, 3, 11$ and $33$. It can't attain others because it can't be greater than $y+x$. If $y-x=1$, $y+x=2013$. But, $AB+AC$ should be more than $BC$, which isn't the case. So, $y-x=1$ is rejected. Similarly, $3$ and $11$ won't be acceptable either. Hence, $y-x=33implies BC=x+y=61$.
answered Jan 2 at 15:10
Ankit KumarAnkit Kumar
1,389219
$endgroup$
Note that $ABX$ and $AXC$ will be right-angled triangles, with $angle AXB$ and $angle AXC$ being $90$ degree. Let $BX=x, CX=y$ and $AX=z$. Then,
$$y^2+z^2=97^2\and\x^2+z^2=86^2$$
Subtract them to get
$$(y+x)(y-x)=97^2-86^2=11cdot183=1cdot3cdot11cdot61$$
$y-x$ can attain the values $1, 3, 11$ and $33$. It can't attain others because it can't be greater than $y+x$. If $y-x=1$, $y+x=2013$. But, $AB+AC$ should be more than $BC$, which isn't the case. So, $y-x=1$ is rejected. Similarly, $3$ and $11$ won't be acceptable either. Hence, $y-x=33implies BC=x+y=61$.
answered Jan 2 at 15:10
Ankit KumarAnkit Kumar
1,389219
answered Jan 2 at 15:10
Ankit KumarAnkit Kumar
1,389219
answered Jan 2 at 15:10
Ankit KumarAnkit Kumar
1,389219
answered Jan 2 at 15:10
Ankit KumarAnkit Kumar
1,389219
1,389219
add a comment |
add a comment |
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$begingroup$
Maybe you can draw a picture and see what you you can deduce after that.
$endgroup$
– Matti P.
Jan 2 at 13:36